Balancing Redox Equations: Half-reaction Method ic solution Acid In acidic solut

Question

Balancing Redox Equations: Half-reaction Method ic solution Acid In acidic solution, the iodate ion can be used to react with a number of metal ions. One such reaction is Since this reaction takes place in acidic solution, H2O(l) and H (aq) will be involved in the reaction. Places for these species are indicated by the blanks in the following In addition to mass balance, oxidation-reduction reactions must be balanced such that the number of electrons lost in the oxidation equals the number of electrons gained in the reduction. This balancing can be done by two methods: the half-reaction method or the oxidation number method. The half-reaction method balances the electrons lost in the oxidation half-reaction with the electrons gained in the reduction half- reaction. In either method H2O() OH (a), and H (aq) may be added to complete the mass balance. Which substances are used depends on the reaction conditions. in the oxidast be 103 (aq) + Sn2+ (aq)1 (aq) + Sn"(aq) restatement of the equation 103 (aq) + Sn2+ (aq) + 1 (aq) + Sn® (aq) +- mass balanc e. Which substances are us Part A What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(1) and H (a) in the appropriate blanks. Your answer should have six terms Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes. Submit Hints My Answers Give Up Review Part

Explanation / Answer

Part A:

In this reaction Oxidation Number (O.N) of I in IO3- is +5 and changes to I-. Thus number of electrons involved in this transformation is 6. i.e., iodine gaines 6 electrons during the reaction.

Sn +2 lost 2 electrons to form Sn+4. Thus the balanced redox reaction without spectator ions as follows:

2IO3-1 + 6Sn+2 + _____ --> 2I-1 + 6Sn+4 + ___.

Since this reaction happened in acidic medium ..

2IO3-1 + 6Sn+2 +12H+ --> 2I-1 + 6Sn+4 + 6H2O

Answer: IO3-1 + 3Sn+2 +6H+ --> I-1 + 3Sn+4 + 3H2O

Part B:

In this reaction, O.N. of Mn in MnO4- is +7 and changes to Mn+4. That is 3 electrons were gained by sulphur

O.N. of S in SO3-2 is +4 changes to +6. That is 2 electrons were lost by suphur

Thus the balanced redox reaction without spectator ions as follows:

2MnO4- + 3SO3-2 + ___ --> 2MnO2 + 3SO4-2 + ___

Since this reaction happened in basic medium

2MnO4- + 3SO3-2 + H2O --> 2MnO2 + 3SO4-2 + 2OH-

Answer: 2MnO4- + 3SO3-2 + H2O--> 2MnO2 + 3SO4-2 + 2OH-

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